dv dy μ [Pa s or kg/ms] - Dynamic viscosity ν = μ/ρ [m 2 /s] - Kinematic viscosity

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2 Modulus of elasticity: E = -dp/(dv/v 1 ) [Pa] Newton s viscosity law shear force: F V τ = = μ = A h dv μ dy μ [Pa s or kg/ms] - Dynamic viscosity ν = μ/ρ [m 2 /s] - Kinematic viscosity Torque (rotational force), T = F r (P = T ω) Angular velocity ω = V/r r = radius, P = pressure

3 Weight = mass gravity acceleration (W = mg, [N = kg m/s 2 ]) (Eqn. 1.4) Weight density (or specific weight)= density gravity acceleration (w = ρg, [N/m 3 = kg /(m 2 s 2 )]) (Eqn. 1.6) (Note w = γ in exercise book) Specific volume ν = reciprocal of density (ν = 1/ρ, [m 3 /kg]) Relative density (or specific gravity), s, is the density normalized with the density of water at a specific temperature and pressure (normally 4 C and atmospheric pressure): s = R.d. = ρ/ρ water (often = ρ/1000) (Eqn. 1.7)

4 Surface tension, σ [N/m], Capillary rise in small tube: h = (2 σ cosθ)/ (ρ g r) Pressure from water column: P = wy = ρgy Force from pressure: F = P A wy = ρgy A P absolute = P atmospheric + P gage

5 For spherical droplet Balance between internal pressure force and surface tension force: p πr 2 = 2πR σ p = 2 σ /R For bubble Balance between internal pressure force and surface tension force: p πr 2 =2 2πR σ p = 4 σ /R

6 FORCE ON SUBMERGED INCLINED PLANE SURFACE (for vertical surface exchange L with h) L P L G L A Resultant force: F= wh G A= ρgh G A Point of action of resultant force: L P = I G /(A L G ) + L G ( Eq. 1.13) L p P=ρgh A: area of plane surface; h G : vertical distance liquid surface - area center; L P : distance O - pressure center; L G : distance O area center; I G : second moment of area about area center axis; L = h/sinθ

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8 Pressure on curved surfaces

9 The resultant force F from vertical and horisontal components is given by: F = F 2 + V F 2 H Direction of the resultant force is given by: tan φ = F V F H Buoyance of body submerged in water: F B = wv (Eq. 1.14) (= ρ g V where V is volume)

10 The general relation for pressure in a static fluid: dp dz = w γ = ρ g => dp = -w dz For a fluid with constant density: z p 1 p 2 = w (z 2 z 1 ) = w h

11 Reynolds number (relationship between velocity and viscosity forces) VD R e= = ν ρvd μ For pipe flow: Laminar flow Re < 2000 Transitional flow Re = 2000 to 4000 Turbulent flow Re > 4000 ν= μ/ρ, V=Q/A ν = kinematic viscosity μ = dynamic viscosity D = diameter (for pipes) Critical Reynolds number, Rc, defining the division between laminar and turbulent flow Parallel walls: Rc 1000 (using mean velocity V and spacing D) Wide open channel: Rc 500 (using mean velocity V and depth D) Flow about sphere: Rc 1 (using approach velocity V and sphere diameter D)

12 Flow types Steady uniform flow: flowrate (Q) and section area (A) are constant Steady non-uniform flow: Q = constant, A = A(x). Unsteady uniform flow: Q = Q(t), A = constant Unsteady non-uniform flow: Q = Q(t), A = A(x). Steady = time independent Uniform = constant section

13 CONTINUITY EQUATION Steady flow ρ 1 V 1 A 1 = ρ 2 V 2 A 2 (m 1 = m 2 ) m 1 m 2 Incompressible flow V 1 A 1 = V 2 A 2 or Q 1 = Q 2 (Q = V A) Control volume Fluid system volume V: Average velocity at a section (m/s) A: Cross-section area (m 2 ) Q: Flow rate (m 3 /s)

14 Equation for a jet trajectory

15 BERNOULLI S EQUATION 2 p V + w γ 2g + z = H = const. Bernoulli s equation is a relationship between pressure, p, velocity, V, and geometric height, z, above a reference plane (datum). Quantity Name Measure of H Energy head Total energy P/w Pressure head Pressure energy Z Elevation head Potential energy V 2 /(2g) Velocity head Kinetic energy H: energy head (m) z: elevation head above datum (m) V: velocity (m/s) g: gravity acceleration (m/s 2 ) p: pressure (Pa) w: weight density for the flowing fluid (N/m 3 ) p + z = piezometric head or wγ H. G. L = Hydraulic Grade Line Trycknivå =

16 ENERGY EQUATION (Bernoulli eqn plus energy added and lost) ( p 1 γw 2 2 V 1 p 2 V 2 + z 1 + ) + h M = ( + z 2 + ) 2g w γ 2g + h L h M h L - energy added by machinery work (pumps, turbines) per unit weight of flowing fluid (m) energy loss per unit weight of flowing fluid (m)

17 POWER IN FLUID FLOW P = w Q h (W) P - Power developed by, for instance, pump or turbine w - weight density (N/m 3 ) Q - h - flow rate (m 3 /s) energy per unit weight of flowing fluid that has been added (h P ) or withdrawn (h T ) Efficiency, η = (power output)/(power input)

18 Momentum equation ΣF = β ρq(v 2 - V 1 ) Correction coefficient for momentum and kinetic energy: β = Σ(v 2 da)/v 2 A (Eqn, 4.17) α = Σ(v 3 da)/v 3 A (Eqn, 4.26)

19 Energy equation for pipe flow: p 1 w γ + where h z 1 h losses L + V 2 V 1 2g = = p 2 wγ h friction + + Pipe flow z V 2 2g h local 16Q 2 2 f = f D 2g or h f = f 5 D 2 2gπ L + h losses General friction formula or Darcy-Weisbach formula (turbulent and laminar flow): h f energy loss due to friction over a distance, L (m), along the pipe f pipe friction factor [f=f(re, Pipe wall roughness ); Fig Moody diagram, laminar flow f = 64/Re; Re = VD/ν] D Pipe diameter (m) V average velocity in the pipe (m/s) Q flowrate in the pipe (m3/s)

20 Local energy losses in pipe flow: h local = K local V 2 2g where h local = local energy loss K local = local loss coefficient (different for different geometry) V 2 /(2g) = kinetic energy (hastighetshöjd)

21 LOCAL ENERGY LOSS ENLARGEMENT ENERGY LOSS FOR OUTFLOW IN RESERVOIR Loss coefficient, K L, for sudden enlargement (V=V 1 ): D 2 /D K L

22 LOCAL ENERGY LOSS - CONTRACTION Loss coefficient for sudden contraction (Franzini and Finnemore, 1997, V = V 2 ): D 2 /D K L

23 Head loss at smooth pipe bends Head loss coefficient for different types of pipe entrances

24 Loss coefficients at right angle bends

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26 Pipe systems branched pipe systems Solution 3 Possible flow situations: 1) From reservoir 1 and 2 to reservoir 3 2) From reservoir 1 to reservoir 2 and 3 3) From reservoir 1 to 3 (Q 2 = 0) For the situation as shown: Energy equation H J = P J /w + z J + V 2 J /2g h f1 + Σh local,1 = z 1 H J h f2 + Σh local,2 = z 2 H J h f3 + Σh local,3 = H J z 3 Continuity equation Q 3 = Q 1 + Q 2 As H J (H J is total head at J) is initially unknown, a method of solution is as follows: 1) Guess H J 2) Calculate Q 1, Q 2, and Q 3 3) If Q 1 + Q 2 = Q 3, then the solution is correct 4) If Q 1 + Q 2 Q 3, then return to 1).

27 NON-STATIONARY PIPE FLOW - OUTFLOW FROM RESERVOIR UNDER VARYING PRESSURE LEVEL When water flows under varying pressure levels the outflow from the reservoir will vary accordingly. In the figure V represents the volume in the reservoir at a certain time. There is also an inflow, Q i, to and an outflow, Q o, from the reservoir.

28 NON-STATIONARY PIPE FLOW, cont. Volume change in the reservoir during a small time interval dt, can be expressed as: dv = (Q i Q o ) dt and dv = A s dz A s dz = (Q i Q o ) dt time. where both inflow and outflow can vary in The outflow can normally be determined by the energy equation that gives outflow as a function of z. For example outflow through a hole: Q o = A hole C D (2gz) 1/2 (Eqn. 5.12) If time is to be estimated that changes the water level from z 1 to z 2 integration of dt = (A s /(Q i Q o )) dz gives: t = z1 z2 (A s /(Q i Q o )) dz this expression can be derivated if Q i = 0 or if Q i = constant and Q o can be re-written as a function of z. Q o can be determined by the energy equation during short time periods assuming stationarity. If water level changes quickly an acceleration term has to be included though.

29 32 μ L h f = 2 w D 2 τ 0 L h f = w R r τ = τ 0 R V (Laminar flow: Poiseuille s equation, 6.11) (Laminar AND turbulent flow)

30 Models for turbulent shear stress τ Boussinesq s model τ = ε (du/dy) where ε is eddy viscosity (laminar τ = μ(du/dy)) Prandtl s model (for pipe flow) τ = ρ l 2 (du/dy) 2 where l is mixing length and function of y, close to wall l = κ y where κ = turbulence constant (~0.4) Equaling the two models gives ε = ρ l 2 (du/dy) (Pa s)

31 Friction in laminar pipe flow Poiseuille s equation: h f = 32νLV / gd 2 (Eq. 6.11) Darcy-Weisbach equation: h f = flv 2 / 2gD (Eq. 6.12) (also for turbulent pipe flow) Equaling gives (laminar flow) 32 ν LV / gd 2 = f LV 2 / 2gD (Re = DV/ ν) f= 64/Re (Eq. 6.13)

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33 Thickness of laminar sublayer Laminar sublayer thickness, δ L (from measurements): 4ν δ L = v * where ν = kinematic viscosity Friction velocity, v *, can be determined from: 2 4τ L L V τ0 h f = = f v* = = V ρgd D 2g ρ 0 f 8 where τ = τ 0 r/r R = radius r = distance from pipe wall τ 0 = shear stress at pipe wall

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35 FORMULAS USED TO DETERMINE THE FRICTION COEFFICIENT Prandtl s and von Karman s semi-empirical laws For smooth pipes 1 Re f = 2log( ) Eq f 2.51 For rough pipes 1 f = 3.7 2log( ) k / D S Eq Colebrook-White transition formula: 1 f k = 2log( 3.7D S Re f ) Eq This formula is applicable to the whole turbulent region for commercial pipes (iterative solution)

36 Moody s diagram

37 NON-CIRCULAR PIPES THE HYDRAULIC RADIUS Head loss calculations for non-circular pipe sections are done using the hydraulic radius concept. The hydraulic radius, R h, is defined as the area, A, of the pipe section divided by the wetted perimeter (circumference), P. R h = A/P For a circular pipe: R h = area/wet perimeter = (πd 2 /4)/(πD) = D/4; or D = 4R h. This value for the diameter may be substituted into the Darcy- Weisbach equation, Reynold s number, and relative roughness: h f = f L 4R H V 2, 2g ρv (4R Re = μ Using these equations, the head loss for non-circular cross-sections may be determined, using an equivalent diameter for a non-circular pipe. The hydraulic radius approach works well for turbulent flow, but not for laminar flow. ) h, k s D k = s 4R h

38 Rotating movement by centrifugal impeller or propeller Pressure increases over the pump p in p out Pressure increases over pump: (p out p in ) / ρg = H p = pump head Specific energy consumption pump = (w Q H p )/(3600 Q η) kwh/m 3 η = power output/power input

39 Energy line h valve Energy line H p = function(qp)

40 A pump is characterized by the pump performance curve

41 Calculation of flowrate and pressure in a pump system Energy line Water is pumped from reservoir A to reservoir B What will the flowrate, Q p, be if the pipe characteristics, L, D, k s are known as well as static head, Δz, and pump curve? The pressure increase over pump (energy supply from pump to water) should achieve two things with respect to lifting water from reservoir A to B: 1) overcome geometric height, Δz 2) overcome head losses h f1 + h f2

42 The hydraulic characteristics for the pipe system, H syst, is obtained from the energy equation L Q 2 H syst = Δz +Σh = Δz + f losses D 2gA 2 (local losses neglected in this case) Pressure increase over pump H syst states how much energy that is needed to transport 1 kg of water from A to B Q p in pipe H p states how much energy the pump can provide to the water When the pump is introduced in the pipe system the flowrate and pump head will adjust so that H syst = H p

43 PARALLEL PUMPING When pumps are operating in parallel they can be replaced by a fictive equivalent pump with a pump curve obtained by horizontal addition of the single pump s pump curves Equivalent pump 2 pumps in parallel

44 PUMPS IN SERIES Pumps operating in series can be considered as being replaced by a fictive equivalent pump with a pump curve obtained by vertical addition of the single pump s pump curves 2 pumps

45 EXAMPLES OF SYSTEM CURVES 1) Two pumps operating in parallel System curve (independent of number of pumps) One pump runs Two pumps run

46 2) Heat exchange system Heat exchange system

47 3)Increase of natural flow rate without pump: Δz = f L D 2 Q 2gA 2 H syst = Δz + f L D 2 Q 2gA 2 = 0 Pump No pump Q without Q with

48 4) Flow control using a valve H syst = Δz + ( Kvalve + f L D ) 2 Q 2gA 2 h valve Choke k v increase valve

49 5) Time-varying reservoir surface H syst = Δz + f L D 2 Q 2gA 2 (Δz varies)

50 6) Flow regulation using speed-adjustable pumps Pump curves Speed regulated pump N 1 (number of rotation)

51 Channel flow Energy eqn for channel [Z + D + α V 2 /2g ] 1 = [Z + D + α V 2 /2g ] 2 + h f

52 Friction along channel bottom τ 0 τ 0 = ρ g R H S 0 where R H = Hydraulic radius S 0 = Channel bottom slope

53 Laminar / Turbulent Re = V (4R )/ ν where R = A/P, (hydraulic radius) A = Area P = wet perimeter For a channel with rectangular section, water depth D and channel width B: A = D B, P = B + 2D R = y B /(B + 2D) ( D for a wide channel) For laminar channel flow: Re < 500

54 Laminar flow, cont. Wide channel R D From above: τ 0 = ρ g R S 0 (1) Linear shear force distribution: τ = τ 0 (1-y/D) (2) Shear force in Newtonian fluid: τ = µ (dv/dy) (3) -where y = vertical coordinate

55 Laminar flow, cont. Eqn 1,2,3 gives: dv/dy = τ/µ = (τ 0 /µ) (1-y/D) dv/dy = (ρ g S 0 /µ) (D-y) v(y) = (ρ g S 0 /µ) (D y-y 2 /2) + C Boundary condition: V (y=0) = 0 C = 0 v(y) = (ρ g S 0 /µ) (D y-y 2 /2) By integration it gives mean velocity V = (g S 0 D 2 ) / (3 ν) where ν = kinematic viscosity

56 Uniform / Non-uniform Uniform flow: d/dx = 0 No change in flow direction (flow, depth) For uniform flow: S o = S f i.e., Bottom parallel with energy line (and water surface)

57 FRICTION and FLOW EQUATIONS From above: τ = ρ g R So (1) Compare general expression τ = K ρ V 2 (2) (1) o (2) V = (g R So / K ) 1/2 Or: V = C ( R So) 1/2 where C = Chezy s coefficient This is Chezy s equation

58 MANNINGS EQUATION (uniform and raw turbulent flow) Manning found: C = R 1/6 / n With Chezy s equation Mannings eqn: V = 1/n R 2/3 S 1/2 o Or: Q = 1/n A R 2/3 S 1/2 o

59 MANNINGS EQUATION, cont.

60 Specific energy, E Specific energy, E, is important for channel flow. E = distance between channel bottom and energy line (D = water depth): E = D + V 2 /(2g) = D + Q 2 /(2g A 2 ) (Eq ) Rectangular and wide channels are emphasized. Flow per width meter, q, is: q = Q/b V = Q/(D b) = q/d E = D + q 2 /(2g D 2 ) q = D (2g (E - D))

61 Critical depth, D c Critical depth is calculated by differentiating above equation (8.22) and solving D subject to a minimum E: q = (g D 3 c) V = (g D c ) D c = 3 (q 2 /g) D c =V 2 /g E min = D c + V 2 c/2g = 3D c /2

62 Froude s number, Fr Fr, Froude s dimensionsless number: Fr = V/( (gd c )) = 1 Fr = 1 critical flow Fr = <1 subcritical flow (small waves travelling upstream) Fr = >1 supercritical flow (small waves do not travel upstream)

63 Critical bottom slope, S c Equation for critical depth is combined with Manning s equation: S c = n 2 g/ 3 D c = n 2 g 10/9 /q 2/9 S 0 >S c S 0 <S c supercritical flow subcritical flow D< D c supercritical flow D> D c subcritical flow

64 Hydraulic jump Momentum eqn Note! Relation symmetrical

65 Hydraulic jump energy relations

66 Water surface profiles - All possible types in Fig Classification from bottom slope - Water depth gradient can be determined from Eqn. 8.41: dd/dl = (S o S f ) / (1 F 2 )

67 Profile types / mild bottom slope

68 Profile types / steep bottom slope

69 Profile types / mild bottom slope, steep bottom slope

70 Control sections examples

71 Step calculation Calculation of water surface is based on energy eqn: [Z + D + V 2 /2g ] 1 = [Z + D + V 2 /2g ] 2 + h f (1) Energy loss h f is calculated as h f = S f L (2) where L = distance between section 1 and section 2 and S f (= energy line slope) from Manning s eqn V = 1/n R 2/3 S f ½ S f = V n / R 4/3 (3) Note that S f is not bottom slope.

72 Calculation of friction slope S f At step calculation of water surface S f can not be expressed exactly since relations vary between section 1 and 2 Some kind of mean value must be used: Alt. 1 S f = (S f1 + S f2 ) / 2 where S f1 = energy line slope in section 1 S f2 = energy line slope in section 2 Alt. 2 S f = V av n / R av 4/3 where V av = (V 1 + V 2 ) / 2 DR av = (R 1 + R 2 ) / 2

73 Direct step calculation Direct step method : - known relationships in section 1 - assume a depth in section 2 - use Manning eqn S f -calculateδl with: ΔL= (E 1 E 2 ) / (S f S o ) ΔL Presupposes: prismatic channel, i.e., geometry known at section 2

74 Sharp-crested weirs Energy eqn (ideal flow) + integration Q = (2/3) (2g) 1/2 B H 3/2 Non-ideal flow compensate (multiply) with discharge coefficient

75 Wide weirs Assume rectangular section Critical conditions Q = g 1/2 B D c 3/2 For practical measurements: Q = C B H 1 3/2

76 Bottom outlets, gates For practical measurements (fig 9.12 a): Q = C a o (2g H 1 ) 1/2 where a o = opening area

77 Flow around bodies Both drag and lift force components can be calculated by: F = C ρ A V 2 /2 C = C lift or C drag (dimensionless) A = characteristic area V = relative speed between body and fluid

78 Boundary layer / surface resistance, cont. Thickness of the laminar boundary layer (δ) grows as δ Re x -1/2 where Re x (local Reynold s number) = V x ρ/μ x = distance from upstream edge At some point downstream the boundary layer becomes turbulent. This occurs approximately at Re x =

79 Boundary layer / surface resistance, cont. Thickness of the turbulent boundary layer (δ) grows as δ Re x -1/5 C f = C f (Re). see Fig. 9.7

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81 Shape resistance, separation Pressure resistance can be calculated as the total integrated pressure (component along flow direction) over the body. At upstream stagnation point the pressure is p = ρ V 2 /2

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83 Flow resistance 3D, cont. For spheres at Re<0.5, C d = 24/Re lead to Stoke s law For larger Reynold s number C d can reduce if the boundary layer becomes turbulent, as this tends to move the separation-point downstream

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